What is the value of $\dfrac{d}{dx}(2x^4+x^3+3x^2)$ at $x=-1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-11$ (Choice B) B $-4$ (Choice C) C $4$ (Choice D) D $-3$
Explanation: Let's first find the expression for $\dfrac{d}{dx}(2x^4+x^3+3x^2)$ and then evaluate it at $x=-1$. According to the sum rule, the derivative of $2x^4+x^3+3x^2$ is the sum of the derivatives of $2x^4$, $x^3$, and $3x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(2x^4)&=2\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=2\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=8x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(2x^4+x^3+3x^2) \\\\ &=2\dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^3)+3\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=2\cdot 4x^3+3x^2+3\cdot 2x&&\gray{\text{The power rule}} \\\\ &=8x^3+3x^2+6x \end{aligned}$ So we found that $\dfrac{d}{dx}(2x^4+x^3+3x^2)=8x^3+3x^2+6x$. Plugging in $x=-1$ and evaluating using the calculator, we find that the value is $-11$. In conclusion, the value of $\dfrac{d}{dx}(2x^4+x^3+3x^2)$ at $x=-1$ is $-11$.